Question

the product of two 2-digit numbers is 833 . if the product of their units digit is 63 and tens digits is 4 , then find the numbers ?​

Answer 1

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A two digit number is such that product of its digit is 18 When 63 is subtracted from the number the digit interchanged their place Find the number

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Solution

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Correct option is A)

Let the number be 10x+y

According to question,

10x+y−63=10y+x

⇒10x−x+y−10y=63

⇒9x−9y=63

⇒x−y=7

⇒x=7+y (i)

xy=18 (ii)

Substituting the value of x in (ii) we get,

(7+y)y=18

⇒y

2

+7y−18=0

⇒y

2

+9y−2y−18=0

⇒y(y+9)−2(y+9)=0

⇒(y+9)(y−2)=0

⇒y=−9 and y=2

y=−9 is not valid

∴y=2

Putting the value of y in (i) we get,

x−2=7

⇒x=7+2

⇒x=9

So the number =10x+y=10(9)+2=92

Answer 2

Answer:

The numbers asked are $49, 17$.

Step-by-step explanation:

Step-1:

Let the two numbers be $xy$ and $ab$

It is given that the product of these two numbers $(xy)*(ab)=833$

The product of their units digit is $y*b=63$

This can only be possible in the following case if one of the units digit is $9$ and the other is $7$.

Step-2:

It is also given that the product of their tens digit is $4$.

This is possible in $2$ cases: $1*4$, $2*2$.

Let us assume the case of $2*2$ then one of the tens digit is $2$ and the other is $2$.

The only numbers possible will be $29, and 27$

$29*27=783\neq 833$

Hence, it is wrong

Let us assume the case of $1*4$, then one of the tens digit is $1$ and the other is $4$.

There will be 4 different cases over here of different combinations of two-digit numbers.

Formulating the cases closely, the best numbers to choose are $49, 17$

$49*17=833$.

Hence, the numbers asked are $49, 17$.

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